Chapter 10: Problem 55
Exercises \(53-56\) give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes. $$ y^{2}-x^{2}=1, \quad \text { left } 1, \text { down } 1 $$
Short Answer
Step by step solution
Identify the Original Hyperbola
Determine the Center Shift
Write the Equation of the New Hyperbola
Identify the New Center
Find the Vertices
Calculate the Foci
Find the Asymptotes
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conic Sections
Equation Transformation
- Horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
- Vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)
To transform it, we perform a shift on the coordinates. This involves substituting \( x \) with \( (x+1) \) and \( y \) with \( (y+1) \). After completing this substitution, the new equation becomes \((y+1)^2 - (x+1)^2 = 1\).
Successfully transforming a hyperbola's equation lays the groundwork for finding its other features post-shift.
Coordinate Shifts
Thus, the original center at \((0,0)\) becomes \((-1, -1)\). This change is executed by substituting \( (x, y) \) with \( (x+1, y+1) \) in the equation. When you change the equation's terms like this, you effectively shift the entire graph without altering the hyperbola's inherent shape.
Understanding this concept is key for redrawing or recalculating a hyperbola's other properties after a shift. Such transformations are crucial for positioning the hyperbola according to a problem's needs or data.
Vertices and Foci of Hyperbolas
In the original equation \( y^2 - x^2 = 1 \), the vertices were at \( (0, \pm1) \). After applying our coordinate shift to \((-1, -1)\), the vertices shift to \((-1, 0)\) and \((-1, -2)\). The foci need to be recalculated with the distance formula, using \( c = \sqrt{a^2 + b^2} \). Since \( a^2 = 1 \) and \( b^2 = 1 \), we compute \( c = \sqrt{2} \).
Thus, the foci of the shifted hyperbola move to \( (-1, -1 \pm \sqrt{2}) \). Calculating these correctly ensures one can fully draft the hyperbola's structure in its new position. Asymptotes such as \( y = \pm x - 1 \), while different from vertices and foci, also provide critical information about the hyperbola’s layout and direction.